using a 7809 voltage regulator

  1. Inadequate smoothing (too much ripple) on the wall wart output for the load, causing the 7809 to go out of regulation when it runs out of headroom.

Fix: if your load current is less than 400mA, place an electrolytic cap large enough to keep the wall-wart's output ripple valleys to > 11V from the input terminal of the regulator to ground. If the load current is higher than 400mA, get a new wall wart.

  1. Too large a load for the wall wart to support, causing it to fall below 11V, the 7809's typical dropout voltage.

Fix: Get a new wall wart.

  1. If not filtered, (or not filtered enough) dropout when the voltage out of the XFMR/bridge combo falls below 11V.

Fix: Figure out how large a filter cap you need from:

IdT C = ----- dV

Where C = the required capacitance in farads I = the average output current in amperes dt= the period of the ripple frequency in seconds dV = the permissible ripple voltage in volts

For a full wave rectifier, dT will be 1/120 = 0.0083s and dV will be the output voltage from the bridge minus 11 volts. The output voltage from the bridge will probably be something like the 23 volt RMS out of the transfromer multiplied by 1.414 to get the peak voltage, then from that, 1.4 volts subtracted for the diode drops in the bridge. That comes to

Vout = (VRMS * sqrt2) - 2Vf = (23 * 1.414) - 1.4 ~ 31V

Subtracting 11V from that to get the permissible ripple voltage gets us 20V for dV so, with the exception of the load current, here ya go:

IdT I * 0.0083s C = ----- = ------------- = ??? dV 20V Once you get the capacitance, don't forget to take into consideration the tolerance of the capacitor, its voltage rating, and its allowable ripple current rating at the ambient temperature in which it'll be running.

Just to be on the safe side of everything, increase the capacitance you get by about 50%, get a cap rated at 50VDC, and make sure the ripple current rating is about twice your load current rating.

BTW, it wouldn't hurt to change the output cap to 0.1µF, and connect it and the input caps as close to the regulator package as you can.

--
John Fields
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John Fields
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^^^^^ ^^^^^ above 11V typ.

--
John Fields
Reply to
John Fields

I have a KA7809 voltage regulator. I have a couple of buffering caps from the Vin to ground (1uF and .22 uF) and from Vout (.01uF).

Seems pretty straightforward, yes?

The problem:

if I use an 18V DC 400 mA wall wart as my Vin I read 7.5V at the Vout. if I use a 23V 1.8A AC transformer through a bridge I read 8.2V at the Vout.

Ive tried 2 different makes of regulator with the same results. What would account for this strange drop in Vout?

FWIW I put my meter on a 9V battery and got... indeed... 9V.

Reply to
catfarm

First, check your pinout. That seems to be the most common problem. Assuming you've got a TO-220 package (metal back, hole in copper for heat sink screw) this should be the view from the front, plastic side facing you (view in fixed font or M$ Notepad):

.---------. | | | o | | | | | '---------' | | | | | | | | | | '-o--o--o-' | | | | | | | | | | | | I G O N N U D T

created by Andy´s ASCII-Circuit v1.24.140803 Beta

formatting link

Once you've checked that, you want to put your meter on AC 20V range and read the input for AC. If you've got a high AC component with no load on the 7809, you've probably got a messed up filter cap in the wall wart. You didn't mention the filter cap with the transformer, so I'm assuming you don't have one. You need a good sized filter cap before the regulator to keep the minimum input voltage to the 7809 below 12.5V, otherwise it won't work right. The wall wart (it IS DC, right?) is supposed to have a big cap built in to keep the ripple on the DC output low.

Try checking the wall wart and replacing it if there's a lot of AC on the output.

If you're using the transformer with rectifier, put a 1000 uF 35V cap from the input terminal of the 7809 to GND.

See if these work first. They're the most common problems.

Good luck Chris

Reply to
CFoley1064

"catfarm" schreef in bericht news:9Ywed.303627$D%.121731@attbi_s51...

First get the datasheet of the 7809. That things sometimes contain usefull information :).

As for the wallwart: A DC wallwart uses to have a smoothing capacitor inside. But you may have a less common type. A 1uF smoothing capacitor is much to small for a normal load. 500uF - 1000uF is a common value. Another thing is the load. Some regulators sometimes need some load for good regulation. Look what happens if you connect a resistor of let's say 1k to the output.

As for the transformer/bridge combination: You sure need to add a smoothing capacitor.

petrus bitbyter.

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Reply to
petrus bitbyter

Well, the first thing you can do is hook the 7809 input to a 12 battery and see if it works. If it does, you have filtering problem with the wall wart. Try a larger capacitor across the input. The size of the capacitor depends on the load current. It's about 8000uF per amp, per volt. So, if the wall wart produces 18 volts rms, the peak will be 1.414 times more, or 25 volts. The 7809 needs a minimum of about 12 volts to deliver 9, so the wall wart voltage can fall from 25 to 12 (13 volts total). if the load current is 500mA, the capacitor will be (8000/13)/2 = 307uF minimum. You can use a larger value for more current.

-Bill

Reply to
Bill Bowden

How did I write that??? Jeez -- I must have gremlins in the keyboard.

Let's see -- I added 9 + 2.5 and got 12.5, and then used below instead of above?? Senior moment. You're right, of course.

Thanks. Chris

Reply to
CFoley1064

Thank you to everyone who responded. I put a 1000uF el cap in the line and everything straightened itself out.

Reply to
catfarm

Just a thought, are the usual small capacitors fitted either side of the regulator? If not it 'could' be oscillating, particularly with the long leads from the wall wart.

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Stan Blazejewski

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